Question

Let \(A\) and \(B\) be disjoint nonempty closed sets in a metric space \(X\), and define $$f(p)=\frac{\rho_{A}(p)}{\rho_{A}(p)+\rho_{\Delta}(p)} \quad(p \in X)$$ Show that \(f\) is a continuous function on \(X\) whose range lies in \([0,1]\), that \(f(p)=0\) precisely on \(A\) and \(f(p)=1\) precisely on \(B .\) This establishes a converse of Exercise 3: Every closed set \(A \subset X\) is \(Z(f)\) for some continuous real \(f\) on \(X\). Setting $$V=f^{-1}\left(\left[0, \frac{1}{2}\right)\right), \quad W=f^{-1}\left(\left(\frac{1}{2}, 1\right]\right)$$ show that \(V\) and \(W\) are open and disjoint, and that \(A \subset V, B \subset W .\) (Thus pairs of disjoint closed sets in a metric space can be covered by pairs of disjoint open sets. This property of metric spaces is called normality.)

Short answer

In summary, we have proven that the function \(f\) is continuous on the metric space \(X\), with a range in \([0,1]\), and that it takes the value 0 on set \(A\) and 1 on set \(B\). Moreover, we have shown that the sets \(V=f^{-1}\left(\left[0, \frac{1}{2}\right)\right)\) and \(W=f^{-1}\left(\left(\frac{1}{2}, 1\right]\right)\) are open, disjoint, and contain the closed sets \(A\) and \(B\) respectively. This demonstrates that metric spaces are normal, which is an important property of topological spaces.

Step-by-step solution

Show f(p) has range in [0,1] #

Since \(\rho_A(p) \ge 0\) and \(\rho_B(p) \ge 0\) for all \(p \in X\), we have that $$0 \le \frac{\rho_A(p)}{\rho_A(p) + \rho_B(p)} \le 1.$$ Thus, the range of \(f(p)\) lies in \([0,1]\).

Show f(p) = 0 on A and f(p) = 1 on B #

For any point \(p \in A\), we have \(\rho_A(p) = 0\) (since \(p\) is in set \(A\)) and \(\rho_B(p) > 0\) (since \(A\) and \(B\) are disjoint). Therefore, $$f(p) = \frac{\rho_A(p)}{\rho_A(p) + \rho_B(p)} = \frac{0}{0 + \rho_B(p)} = 0$$ for all \(p \in A\). Similarly, for any point \(p \in B\), we have \(\rho_A(p) > 0\) and \(\rho_B(p) = 0\) (since \(p\) is in set \(B\)). Therefore, $$f(p) = \frac{\rho_A(p)}{\rho_A(p) + \rho_B(p)} = \frac{\rho_A(p)}{\rho_A(p) + 0} = 1$$ for all \(p \in B\).

Show f(p) is continuous #

Let \(p \in X\). Given \(\epsilon > 0\), we want to find \(\delta > 0\) such that for all \(x \in X\) with \(d(x,p) < \delta\), we have \(|f(x) - f(p)| < \epsilon\). Let \(x \in X\) with \(d(x,p) < \delta\). Observe that $$\left|\frac{\rho_A(p)}{\rho_A(p) + \rho_B(p)} - \frac{\rho_A(x)}{\rho_A(x) + \rho_B(x)}\right| \le \left|\rho_A(p) - \rho_A(x)\right| + \left|\rho_B(p) - \rho_B(x)\right|.$$ By the triangle inequality, \(|\rho_A(p) - \rho_A(x)| \le d(x,p)\) and \(|\rho_B(p) - \rho_B(x)| \le d(x,p)\). Thus, choose \(\delta = \epsilon\). Then, if \(d(x,p) < \delta = \epsilon\), we have $$\left|f(p) - f(x)\right| \le d(x,p) < \epsilon,$$ which shows that \(f\) is continuous in \(X\).

Show V and W are open and disjoint #

To show that \(V\) and \(W\) are open, let \(p \in V\). Then, \(f(p) < \frac{1}{2}\). Since \(f\) is continuous, there exists \(\delta > 0\) such that for all \(x \in X\) with \(d(x,p) < \delta\), \(|f(x) - f(p)| < \frac{1}{2} - f(p)\). Thus, \(f(x) < \frac{1}{2}\), which means \(x \in V\). Therefore, \(V\) is open. Similarly, let \(p \in W\). Then, \(f(p) > \frac{1}{2}\). Since \(f\) is continuous, there exists \(\delta > 0\) such that for all \(x \in X\) with \(d(x,p) < \delta\), \(|f(x) - f(p)| < f(p) - \frac{1}{2}\). Thus, \(f(x) > \frac{1}{2}\), which means \(x \in W\). Therefore, \(W\) is open. Since the range of \(f(p)\) lies in \([0,1]\), no point in \(X\) can have \(f(p) = \frac{1}{2}\). Thus, \(V\) and \(W\) are disjoint as the pre-images don't overlap.

Show A is a subset of V and B is a subset of W #

For any \(p \in A\), we have \(f(p) = 0\). Therefore, \(p \in f^{-1}\left(\left[0, \frac{1}{2}\right)\right) = V\), and thus \(A \subseteq V\). For any \(p \in B\), we have \(f(p) = 1\). Therefore, \(p \in f^{-1}\left(\left(\frac{1}{2}, 1\right]\right) = W\), and thus \(B \subseteq W\). This completes the proof that the given function \(f\) satisfies all the required properties, and that the sets \(V\) and \(W\) are open and disjoint, containing the closed sets \(A\) and \(B\) respectively, which demonstrates that metric spaces are normal.

Key concepts

Continuous Functions

A function is continuous if small changes in the input lead to small changes in the output. In a metric space, this means that for any point \(p\) in the space and any small number \(\epsilon > 0\), there exists a small number \(\delta > 0\) such that if a point \(x\) is within \(\delta\) of \(p\), then the value of the function at \(x\) is within \(\epsilon\) of the function's value at \(p\). This concept ensures the output of the function does not have any abrupt jumps or breaks.

The function \(f(p) = \frac{\rho_A(p)}{\rho_A(p) + \rho_B(p)}\) was shown to be continuous by examining how it behaves as \(x\) approaches \(p\). By picking \(\delta = \epsilon\), we know that whenever the distance between \(x\) and \(p\) is less than \(\delta\), the difference between \(f(x)\) and \(f(p)\) will be less than \(\epsilon\). This property confirms the function is smooth over the entire metric space \(X\).

Closed Sets

In metric spaces, a closed set is one that includes all its boundary points. If you have a point that's not in the set but can get arbitrarily close to it, it is considered a boundary point, and in a closed set, such points are part of the set.

For the exercise, both sets \(A\) and \(B\) are closed. This characteristic is crucial because it guarantees that each set includes all points where we might consider switching from one set to the other. This closure property ensures that their distances, \(\rho_A\) and \(\rho_B\), highlight every potential change as \(p\) moves through space. It also means that the points at which \(f(p)\) equals 0 or 1 due to \(\rho_A(p)\) or \(\rho_B(p)\) being zero are all inside \(A\) or \(B\), allowing the function to behave exactly as predicted at those boundaries.

Open Sets

Open sets in metric spaces are collections of points where, for any point within the set, you can find a small "bubble" around it that is entirely contained within the set. This means none of the boundary points are included in an open set.

In the given problem, the sets \(V = f^{-1}([0, \frac{1}{2}))\) and \(W = f^{-1}((\frac{1}{2}, 1])\) are open. For any point \(p\) in \(V\), because the function \(f\) is continuous, you can find a neighborhood around \(p\) that is also within \(V\). The same applies for \(W\).

• This implies that if a point is slightly perturbed, it remains in either \(V\) or \(W\).
• This ensures \(V\) and \(W\) have no common points, making them disjoint.

Disjoint Sets

Disjoint sets are sets that have no elements in common. When sets \(A\) and \(B\) do not overlap at any point, they are considered disjoint.

In the solution, \(A\) and \(B\) were initially disjoint closed sets. The function \(f\) helps us create two new open sets, \(V\) and \(W\), which also do not overlap. This separation signifies they are disjoint.

• For a point in \(V\), the value of \(f\) is always less than 0.5.
• For a point in \(W\), the value of \(f\) is always greater than 0.5.

Since no point can make \(f(p) = 0.5\), it guarantees no overlap between \(V\) and \(W\). Hence, metric spaces allow such separate open coverages for closed disjoint sets.

Normality of Metric Spaces

The normality in metric spaces is an essential property stating that any two disjoint closed sets can be separated by two disjoint open sets. This ensures that even if sets seem entirely separate, the space's structure can ensure they are isolated.

In this exercise, the normality is demonstrated by the set function \(f\) creating \(V\) and \(W\) that contain \(A\) and \(B\) respectively without intersecting. This property of metric spaces is valuable because:

• It allows closed sets to be "spread apart" using open sets, providing flexibility in analysis.
• Supporting the construction of continuous functions that can separate points.

The normality assures us of the ability to manage separate regions within metric spaces effectively, thus, highlighting its importance in complex mathematical and real-world applications.